I'm a math PhD student, part of my requirement is that I have to teach classes. So I'm often the "prof" (instructor) of the class. Sometimes I wear Veil of Maya, Safety Fire, Ever Forthright, etc to class, but in my whole first year of teaching I haven't gotten a single comment. :(

WOW, I feel for you ... it must be cathartic to live on that. I really wanted to go to University of Colorado, and it all checked out except the money. International tuition is insane. At my school now I have to do 2/2, but only as an assistant --- marking homework, weekly tutorials ... I don't have to run a lecture.

Thanks, that's really encouraging. Hmm, I thought I'd need more than $100/mo to keep a dog ... But googling it says ~$1000/yr, so I guess that sounds right. I'm mostly worried about the financial aspects of keeping a dog, rather than the schedule.

What are the cons of leaving your dog at home during the day? I imagine there's the risk of having your rugs and cushions ripped up, or pooping on the floor ... is there anything else? If I had a dog I'd try to stay home as much as possible, like I'd only go to school for meetings/classes. It takes me about 25 minutes to walk to school anyway.

Also, I live on the 9th floor of an apartment building ... Would I have to get him an indoor potty instead of letting him go outside?

Congrats on the student loans! My situation is similar to yours. I have lots of student loans, but I'm a PhD student and I still have another four years before I have to start paying them off. My stipend is about $28,000/yr and my rent is only $360/mo (I live with two other people). I cook a lot, so groceries only cost me ~$250 a month. I like being by myself and I'm easily entertained, so I pretty much only spend entertainment money on video games. I also play guitar, which doesn't cost me very much (just $10 to restring every month or so).

Here's my question: I'd like to get a dog. Do you think I could pull it off?

I always thought you'd need at least some experience in finance to become an actuary.

I have a BMath specializing in pure math and I'm just finishing my first year of grad school in pure math. When I'm done I'll have a thesis in pure math but no finance skills. Would it be impossible for me to get a job as (say) an actuary?

It's very remarkable that geometric properties of algebraic curves can be encoded using a totally algebraic object.

Here's how it works. Given an algebraic curve X and a point p on it, you can define a ring O_p, pretty much it consists of functions defined on an open neighborhood near p. It's called the local ring of X at p. This has a special ideal m, which consists of those functions which vanish at p. In fact m is the unique maximal ideal of O_p. It turns out that the following are equivalent:

O_p is a dvr,

m is a principal ideal (this is just ring theory), and

X is smooth at p (meaning that X has nonzero partial derivatives at p).

I was dating a girl for a few months and somehow music never came up, and then I found out she listened to Daughters, Converge, old Norma Jean, even Number Twelve. But she looked like the type of girl who only listened to Taylor Swift. No idea how we avoided noticing that common interest for so long, it was awesome.

Lemme try to put a spin on it that's different from representations of groups: instead, representations of associative rings. These are basically just R-modules. It's called representation theory because an R-module is a way to represent R as a ring of endomorphisms of an abelian group. Sometimes, viewing R acting on a group (external information) can tell you more about R (internal information), e.g. ideals. For example, the kernel of a simple (= irreducible) representation is called a primitive ideal, and these are closely related to prime ideals. Both these types of ideals give you topological invariants of R (Spec R and Prim R).

why might it be interesting/relevant for a student generally more interested in Analysis?

There are many special types of rings/algebras in analysis --- namely, operator algebras, C-algebras, and group algebras. These can be studied by their representations, the same way rings can be studied by their modules. For example, there is a deep theorem in C-algebras that every C-algebra A has a *faithful representation on a Hilbert space H. This means an injective *-homomorphism from A to the space B(H) of bounded operators on H --- therefore A can be realized as a self-adjoint operator algebra.

Representation theory is also very important in the study of topological groups. I know much less about that though. In Lie groups, representation theory is used to classify the simple Lie algebras.

It's impossible to find such matrices, as everyone else has pointed out (in particular /u/elseifian gave a great proof). But let me explain a far more general fact that gives an unnecessarily abstract proof of the theorem you're asking.

The algebra of matrices is finite-dimensional, hence Noetherian. It's true in any Noetherian ring that if xy = 1 then yx = 1 --- so left-invertible elements are units. A ring with this property is called Dedekind-finite. One reason for this name is that the condition "xy = 1 implies yx = 1" is regarded as a feature of finite-dimensional algebras; in fact, people in operator algebras abbreviate Dedekind-finiteness to just "finite". (The only Noetherian operator algebras are finite-dimensional ones, anyway.)

Here is an even more general fact.

Theorem. Let M be a Noetherian module over a ring R and let f : M → M be a module homomorphism. If f is surjective then f is injective.

Let's see how this entails the claim that Noetherian rings are Dedekind-finite. If R is a Noetherian ring then in particular it is a Noetherian module. Suppose xy = 1 and consider the module homomorphism f : R → R given by f(r) := xr. Then for any element r we get f(yr) = xyr = 1r = r, and therefore f is surjective. By the above theorem, f must be injective. But f(yx - 1) = x(yx - 1) = xyx - x = 0, so injectivity implies yx = 1. QED!

So now let me give a hint of how to prove the main theorem ... given the module homorphism f, let f2, f3, f4, ... be obtained by iterating f, and consider the ascending chain of submodules

ker f ⊆ ker f2 ⊆ ker f3 ⊆ ...

By Noetherian-ness, notice that this has to stabilize, say ker fn = ker fn+1 . ... Finish it off!

That's a good point. I guess I should have said "remaster" rather than "remake". I don't mean a remake on the scale of FFVII, rather just an HD port like they did for Resident Evil, MGS 2/3, Jet Set Radio, etc.